Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 5 Advertisements PROBLEM: Evaluate limx→∞(8x−54x2+3)\displaystyle \lim\limits_{x\to \infty }\left(\frac{8x-5}{\sqrt{4x^2+3}}\right)x→∞lim(4x2+38x−5) Advertisements SOLUTION: Divide by the highest denominator power limx→∞(8x−54x2+3)=limx→∞(8x−54x2+3⋅1x1x)=limx→∞(8xx−5x4x2x2+3x2)=limx→∞(8−5x4+3x2)=8−04+0=82=4 (Answer)\begin{align*} \displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{8x-5}{\sqrt{4x^2+3}}\right) & =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{8x-5}{\sqrt{4x^2+3}}\cdot \displaystyle \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\right) \\ \\ & =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{8x}{x}-\displaystyle \frac{5}{x}}{\sqrt{\displaystyle \frac{4x^2}{x^2}+\displaystyle \frac{3}{x^2}}}\right)\\ \\ & =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{8-\displaystyle \frac{5}{x}}{\displaystyle \sqrt{4+\displaystyle \frac{3}{x^2}}}\right) \\ \\ & =\displaystyle \frac{8-0}{\sqrt{4+0}} \\ \\ & =\displaystyle \frac{8}{2} \\ \\ & =\displaystyle 4 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}x→∞lim(4x2+38x−5)=x→∞lim⎝⎛4x2+38x−5⋅x1x1⎠⎞=x→∞lim⎝⎛x24x2+x23x8x−x5⎠⎞=x→∞lim⎝⎛4+x238−x5⎠⎞=4+08−0=28=4 (Answer) Advertisements Advertisements